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Greeley [361]
2 years ago
14

A signal that cannot be faked and carries the most accurate information about a situation or individual is known as an ______ si

gnal
Engineering
1 answer:
Pie2 years ago
5 0

Answer: Honest

Explanation: In other to establish communication between two or more species, the sender and receiver, It is required of the sender to showcase certain behavior, sound or other demonstrations which are capable of passing a message to the receiver. These demonstrations are often reffered to as signals. Signals could be honest or dishonest. Honest signals are characterized by it's usefulness to the receiver and the conveyance of the actual or true meaning of underlying signal being transmitted. This is the opposite of dishonest signals which are often used to trick the receiving party as the information being transmitted are inaccurate and unreliable.

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A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in
RideAnS [48]

To resolve this problem we have,

R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm

F_{f2} is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa

After that, we can calculate the flexural strenght for the square cross section using the previously value.

\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN

6 0
2 years ago
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the
morpeh [17]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.                                          

<u>Explanation</u>:        

<u>Given</u>:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

<u>To calculate</u>:

The magnitude of applied stress in the direction of [101] and [011].

<u>Formula</u>:

zcr=σ cosФ cosλ

<u>Solution</u>:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ  = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.                                                                                                                                                                      

                                                                                   

                                                                                                                                                   

                                                                                                                                                             

5 0
2 years ago
A cylindrical specimen of a brass alloy having a length of 60 mm (2.36 in.) must elongate only 10.8 mm (0.425 in.) when a tensil
Gemiola [76]

The radius of the specimen is 60 mm

<u>Explanation:</u>

Given-

Length, L = 60 mm

Elongated length, l = 10.8 mm

Load, F = 50,000 N

radius, r = ?

We are supposed to calculate the radius of a cylindrical brass specimen in order to produce an elongation of 10.8 mm when a load of 50,000 N is applied. It is necessary to compute the strain corresponding to this

elongation using Equation:

ε = Δl / l₀

ε = 10.8 / 60

ε = 0.18

We know,

σ = F / A

Where A = πr²

According to the stress-strain curve of brass alloy,

σ = 440 MPa

Thus,

sigma = 50,000 / \pi  (r)^2\\\\440 X 10^6 = \frac{50,000}{3.14 X (r)^2}\\\\r = 0.06m\\r = 60mm\\\\\\

Therefore, the radius of the specimen is 60 mm

3 0
2 years ago
Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is
GarryVolchara [31]

Answer:

Code fixed below using Java

Explanation:

<u>Error.java </u>

import java.util.Random;

public class Error {

   public static void main(String[] args) {

       final int MAXCHEESE = 10;

       String[] names = new String[MAXCHEESE];

       double[] prices = new double[MAXCHEESE];

       double[] amounts = new double[MAXCHEESE];

       // Three Special Cheeses

       names[0] = "Humboldt Fog";

       prices[0] = 25.00;

       names[1] = "Red Hawk";

       prices[1] = 40.50;

       names[2] = "Teleme";

       prices[2] = 17.25;

       System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

       System.out.println(names[0] + ": $" + prices[0] + " per pound");

       System.out.println(names[1] + ": $" + prices[1] + " per pound");

       System.out.println(names[2] + ": $" + prices[2] + " per pound");

       Random ranGen = new Random(100);

       // error at initialising i

       // i should be from 0 to MAXCHEESE value

       for (int i = 0; i < MAXCHEESE; i++) {

           names[i] = "Cheese Type " + (char) ('A' + i);

           prices[i] = ranGen.nextInt(1000) / 100.0;

           amounts[i] = 0;

           System.out.println(names[i] + ": $" + prices[i] + " per pound");

       }        

   }

}

7 0
2 years ago
. A piston-cylinder device whose piston is resting on top of a set of stops initially contains 0.5 kg of helium gas at 100 kPa a
Delvig [45]

Answer:

Qin = 1857 kJ

Explanation:

Given

m = 0.5 Kg

T₁ = 25°C = (25 + 273) K = 298 K

P₁ = 100 kPa

P₂ = 500 kPa

First, the temperature when the piston starts rising is determined from the ideal gas equations at the initial state and at that state:

T₂ = T₁*P₂/P₁

⇒  T₂ = 298 K*(500 kPa/100 kPa) = 1490 K

Until the piston starts rising no work is done so the heat transfer is the change in internal energy

Qin = ΔU = m*cv*(T₂-T₁)

⇒   Qin = 0.5*3.1156*(1490 - 298) kJ = 1857 kJ

8 0
2 years ago
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